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A158909
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Riordan array (1/((1-x)(1-x^2)), x/(1-x)^2). Triangle read by rows, T(n,k) for 0 <= k <= n.
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6
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1, 1, 1, 2, 3, 1, 2, 7, 5, 1, 3, 13, 16, 7, 1, 3, 22, 40, 29, 9, 1, 4, 34, 86, 91, 46, 11, 1, 4, 50, 166, 239, 174, 67, 13, 1, 5, 70, 296, 553, 541, 297, 92, 15, 1, 5, 95, 496, 1163, 1461, 1068, 468, 121, 17, 1, 6, 125, 791, 2269, 3544, 3300, 1912, 695, 154, 19, 1
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OFFSET
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0,4
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COMMENTS
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Diagonal sums are the Jacobsthal numbers A001045.
Transforms r^n into the symmetric third-order sequence with g.f. 1/(1-(r+1)x-(r+1)x^2+x^3), see the formulas.
The signed triangle t(n, k) = (-1)^(n-k)*T(n, k) appears in the expansion [n+2, 2]_q / q^n = Sum_{k=0} t(n, k)*y^(2*k), with y = q^(1/2) + q^(-1/2), where [n+2, 2]_q are q-binomial coefficients (see A008967, but with a different offset). The formula is [n+2, 2]_q / q^n = S(n+1, y)*S(n, y)/y with Chebyshev S polynomials (A049310). This is a polynomial in y^2 but not in q after replacement of the given y = y(q).
The A-sequence for this Riordan triangle is A(n) = (-1)^n*A115141(n) with o.g.f A(x) = 1 + x*(1 + c(-x)), with c(x) generating A000108 (Catalan).
The Z-sequence is z(n) = (-1)^(n+1)*A071724(n), for n >= 1 and z(0) = 1. The o.g.f. is Z(x) = 1 + x*c(-x)^3. See A071724 for a link on A- and Z-sequences, and their use for the recurrence. (End)
T(n,k) is the number of tilings of a (2*n+1)-board (a 1 X (2*n+1) rectangular board) using 2*k+1 squares and 2*(n-k) (1,1)-fences. A (1,1)-fence is a tile composed of two squares separated by a gap of width 1. - Michael A. Allen, Mar 20 2021
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LINKS
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FORMULA
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Sum_{k=0..n} T(n,k) = Fibonacci(n+1)*Fibonacci(n+2) = A001654(n+1).
T(n, k) = Sum_{i=0..n-k} (-1)^(i+n-k) * binomial(i+2*k+1, i).
Sum_{k=0..n} T(n, k)*r^k = coeftayl(1/(1-(r+1)*x-(r+1)*x^2+x^3), x=0, n). [Barry] (End)
T(n, k) = T(n-1, k) + T(n-1, k-1) + T(n-2, k) + T(n-2, k-1) - T(n-3, k), T(0, 0) = 1, T(n, k) = 0 if k<0 or if k>n. - Philippe Deléham, Nov 12 2013
O.g.f. for the row polynomials (that is for the triangle): G(z, x) = 1/((1 + z)*(1 - (x + 2)*z + z^2)), and
O.g.f. for column k: x^k/((1+x)*(1-x)^(2*(k+1)) (Riordan property). (End)
T(n, k) = binomial(k + n + 2, n - k + 1)*hypergeom([1, k + n + 3], [n - k + 2], -1) + (-1)^(n - k)/4^(k + 1). - Peter Luschny, Oct 31 2019
T(n,k) = T(n-2,k) + binomial(n+k,2*k). (End)
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EXAMPLE
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The triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 ...
----------------------------------------------------
0: 1
1: 1 1
2: 2 3 1
3: 2 7 5 1
4: 3 13 16 7 1
5: 3 22 40 29 9 1
6: 4 34 86 91 46 11 1
7: 4 50 166 239 174 67 13 1
8: 5 70 296 553 541 297 92 15 1
9: 5 95 496 1163 1461 1068 468 121 17 1
10: 6 125 791 2269 3544 3300 1912 695 154 19 1
...
----------------------------------------------------------------------------
Recurrence: T(5, 2) = 16 + 13 + 5 + 7 - 1 = 40, and T(5, 0) = 3 + 2 - 2 = 3. [using Philippe Deléham's Nov 12 2013 recurrence]
Recurrence from A-sequence [1, 2, -1, 2, -5, ...]: T(5, 2) = 1*13 + 2*16 - 1*7 + 2*1 = 40.
Recurrence from Z-sequence [1, 1, -3, 9, -28, ...]: T(5, 0) = 1*3 + 1*13 - 3*16 + 9*7 - 28*1 = 3. (End)
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MAPLE
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T := (n, k) -> binomial(k+n+2, n-k+1)*hypergeom([1, k+n+3], [n-k+2], -1) + (-1)^(n-k)/4^(k+1):
seq(seq(simplify(T(n, k)), k=0..n), n=0..9); # Peter Luschny, Oct 31 2019
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MATHEMATICA
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Table[Sum[(-1)^(j+n-k)*Binomial[j+2*k+1, j], {j, 0, n-k}], {n, 0, 12}, {k, 0, n}] // Flatten (* G. C. Greubel, Mar 18 2021 *)
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PROG
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(Sage) flatten([[sum((-1)^(j+n-k)*binomial(j+2*k+1, j) for j in (0..n-k)) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 18 2021
(Magma) [(&+[(-1)^(j+n-k)*Binomial(2*k+j+1, j): j in [0..n-k]]): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 18 2021
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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